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Line Integral over circle region

  1. May 16, 2016 #1
    1. The problem statement, all variables and given/known data
    Evaluate ∫c (x + y) ds, where C is the circle centred at (1/2, 0) with radius 1/2.

    2. Relevant equations


    3. The attempt at a solution
    parametrise
    x=1/2cos(t)
    y=1/2sin(t)
    0≤t≤2π
    ds=√dx2+dy2
    =√(1/2)2-sin2(t)+(1/2)2cos2(t)
    =√-(1)2(1/2)2sin2(t)+(1/2)2cos2(t)
    =√-(1)2(1/2)2(sin2(t)+cos2(t))
    ds=1/2

    ∫ [(1/2)cos(t) + (1/2)sin(t)]*(1/2) dt, where 0≤t≤2π

    I evaluated this integral and got 0. Is this because C is a simple closed curve?
     
  2. jcsd
  3. May 16, 2016 #2

    Ray Vickson

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    I evaluated the (correct) integral and obtained an answer of ##\pi/n##, where ##n## is an integer whose exact value I will not divulge at this time. I assumed the circle was traversed in a counter-clockwise direction, which you did not mention in the problem statement.
     
  4. May 16, 2016 #3
    The problem statement didn't mention the direction. So I just assume anti-clockwise.

    Do you know where I have made my mistake? I have been on this problem for a while now and can't seem to find it. I've put the integral into a computation as well and got back 0 as the answer.
     
  5. May 16, 2016 #4

    Ray Vickson

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    PF rules forbid me from telling you exactly where your mistake lies. All I can do is urge you to be more careful, and to read the question carefully.
     
  6. May 16, 2016 #5
    Argh!

    The circle is centred at (1/2,0) not at the origin. I'm not entirely sure how this changes anything though.
     
  7. May 16, 2016 #6

    SteamKing

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    Since C is a circle, perhaps a more natural expression for ds would be r ⋅ dθ
     
  8. May 16, 2016 #7
    I found a source online that suggested if the circle is not at the origin then:
    x = h + r cos(t)
    y = k + r sin(t)
    so
    x = 1/2 + (1/2 cos(t))
    y = 1/2 sin(t)
    edit: FOUND THE INFO I WAS LOOKING FOR!

    Circle with Center at Point (h,k)
    http://www.regentsprep.org/regents/math/geometry/gcg6/LCirh3.gif
    where the center is (h,k)
    and the radius is r

    so

    x= 1/2cos(t)-1/2
    y= 1/2sin(t)
     
  9. May 16, 2016 #8

    Ray Vickson

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    In one place you wrote ##x = (1/2) + (1/2)\cos(t)##, while in another place you wrote ##x = -(1/2) + (1/2)\cos(t)##. So, which is it?
     
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